1. **State the problem:** We have a curve with equation $$y = x^3 + ax + b$$ and stationary points at coordinates $(2, k)$ and $(-2, 10 - k)$. We need to find the values of $a$, $b$, and $k$. 2. **Recall the formula for stationary points:** Stationary points occur where the derivative of the function equals zero. The derivative of $$y = x^3 + ax + b$$ is $$y' = 3x^2 + a$$. 3. **Find $a$ using the stationary points:** Since $(2, k)$ and $(-2, 10 - k)$ are stationary points, their $x$-coordinates satisfy $$y' = 0$$. Set $$y' = 0$$ at $$x=2$$: $$0 = 3(2)^2 + a = 3 \times 4 + a = 12 + a$$ Solve for $a$: $$a = -12$$ 4. **Use the original function to express $k$ and $b$:** At $x=2$, $$y = k$$, so: $$k = (2)^3 + a(2) + b = 8 + (-12)(2) + b = 8 - 24 + b = b - 16$$ At $x=-2$, $$y = 10 - k$$, so: $$10 - k = (-2)^3 + a(-2) + b = -8 + (-12)(-2) + b = -8 + 24 + b = b + 16$$ 5. **Solve for $k$ and $b$ using the two equations:** From the first: $$k = b - 16$$ From the second: $$10 - k = b + 16$$ Substitute $k$ from the first into the second: $$10 - (b - 16) = b + 16$$ Simplify: $$10 - b + 16 = b + 16$$ $$26 - b = b + 16$$ Add $b$ to both sides: $$26 = 2b + 16$$ Subtract 16: $$26 - 16 = 2b$$ $$10 = 2b$$ Divide both sides by 2: $$\cancel{2}b / \cancel{2} = 10 / 2$$ $$b = 5$$ 6. **Find $k$ using $b=5$:** $$k = b - 16 = 5 - 16 = -11$$ **Final answers:** $$a = -12, \quad b = 5, \quad k = -11$$